Brush question 5

16、 String multiplication

class Solution:
    def multiply(self, num1, num2):
        num1 = num1[::-1]
        num2 = num2[::-1]
        length1 = len(num1)
        length2 = len(num2)
        temp = [0 for __ in range(length1 + length2)] 
        for i in range(len(num1)):
            for j in range(len(num2)):
                temp[i+j] += int(num1[i]) * int(num2[j])
        print temp
        list_ = []
        for i in range(len(temp)):
            digit = temp[i] % 10 
            carry = temp[i] / 10  
            if i < len(temp)-1:
                temp[i+1] += carry
            list_.insert(0, str(digit)) 
        while list_[0] == '0' and len(list_) > 1:
            list_.pop(0)
        return ''.join(list_)

20、 Spiral matrix

class Solution:
    def spiralOrder(self, matrix):
        res=[]
        m=len(matrix)
        if m==0:
            return res
        else:
            n=len(matrix[0])
            if n==0:
                return res
        count=(m+1)//2
        k=0
        su=m*n
        cc=0
        while k<count and cc<su:
            for i in range(k,n-k):
                res.append(matrix[k][i])
                cc+=1
            for i in range(k+1,m-1-k):
                res.append(matrix[i][n-1-k])
                cc+=1
            if k!=m-1-k:
                for i in range(k,n-k):
                    res.append(matrix[m-1-k][n-1-i])
                    cc+=1
            if k!=n-1-k:
                for i in range(k+1,m-1-k):
                    res.append(matrix[m-1-i][k])
                    cc+=1
            k+=1
        return res

21、 Rotate the list

class Solution:

    def rotateRight(self, head: ListNode, k: int) -> ListNode:

        if not head:

            return None

        length = 0

        index = head

        while index.next:

            index = index.next

            length += 1

        index.next = head

        length = length + 1

        k = k % length

        for i in range(length - k):

            head = head.next

            index = index.next

        index.next = None

        return head